LeetCode 746 使用最小花费爬楼梯

动态规划

const int N = 1000;
class Solution {
public:
    int dp[N];
    int minCostClimbingStairs(vector<int>& cost) {
        dp[0] = cost[0];
        dp[1] = cost[1];

        for (int i = 2; i < cost.size(); i ++) //最后一步相当于不花费
            dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i];

        return min(dp[cost.size() - 1], dp[cost.size() - 2]); //最后要么从倒数第二层爬要么从倒数第一层爬
    }
};

动态规划压缩到维护两个值

class Solution {
public:
    int dp[2];
    int minCostClimbingStairs(vector<int>& cost) {
        dp[0] = cost[0];
        dp[1] = cost[1];

        for (int i = 2; i < cost.size(); i ++) {
            //最后一步相当于不花费
            int temp = dp[1];
            dp[1] = min(dp[0], dp[1]) + cost[i];
            dp[0] = temp;
        } 

        return min(dp[0], dp[1]); //最后要么从倒数第二层爬要么从倒数第一层爬
    }
};